Lifting The Exponent

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The Lifting the Exponent (LTE) Lemma is a useful lemma about the largest power of a prime dividing a difference or sum of \(n^\text{th}\) powers. Here are some sample problems whose solutions use the lemma.

Let \( n \) be a squarefree integer. Show that there is no pair of coprime positive integers \( x,y\) such that \[(x+y)^3 | (x^n+y^n).\]

Show that \( 2 \) is a primitive root mod \( 3^k \) for all positive \( k \).

Without LTE, the problem can be solved by factoring
\[
\begin{align}
5^{18}-2^{18}&=(5^9-2^9)(5^9+2^9) \\
&= (5^3-2^3)(2^6+2^35^3+5^6)(5^9+2^9) \\
&= (5-2)(2^2+(2)(5)+5^2)(2^6+2^35^3+5^6)(5^9+2^9)
\end{align}
\]
The first factor has one \( 3 \) and the fourth factor has no \( 3\)s, and some careful mod-\(9\) analysis shows that the second and third factors are divisble by \(3 \) but not \( 9 \), so the total number of factors of \( 3 \) is \(3\). This is quite a bit more complicated (but note that it also indicates how an inductive proof of LTE might proceed).

This lemma gives a practical way to solve many problems involving the largest power of a prime that divides certain expressions. In particular, the solutions to the problems in the introduction all use LTE in an essential way.

Solution to Problem 1

Assume \( (x+y)^3|(x^n+y^n) \) with gcd\((x,y)=1\). We will derive a contradiction.

First suppose \( n \) is even. If there is an odd prime \( p |(x+y) \), then \( x^n+y^n \equiv x^n +(-x)^n \equiv 2x^n \) mod \(p \), so \( p|x \), so \(p|y \), contradiction. Since \( x \) and \( y \) are positive, the only possible way that there is no odd prime \(p\) dividing \( x+y \) is if \( x+y \) is a power of \( 2\). In this case, \(x\) and \( y \) are both odd since they are coprime, so since \(n \) is even \( x^n \) and \( y^n \) are both \( 1 \) mod \( 8 \), so \( v_2(x^n+y^n) = 1 \), but \( v_2((x+y)^3) \ge 3 \), so again we get a contradiction.

As above, the only remaining case is when \( x+y \) is a power of \( 2 \). But in this case, \( v_2(x^n+y^n) = v_2(x+y) \) if \( n \) is odd, because \( x \) and \( y \) are odd and the expansion of
\[
\frac{x^n+y^n}{x+y} = x^{n-1} -x^{n-2}y + \cdots - xy^{n-2}+y^{n-1}
\]
has an odd number of terms, all of which are odd. So it is impossible for \( (x+y)^3 \) to divide \( x^n+y^n \), since the power of \( 2 \) on the left exceeds the power of \(2 \) on the right. \( \square \)

How many pairs of positive integers \((x, y)\) satisfy the equation above?

Solution to Problem 3

If one of \( x \) or \( y \) is divisible by \(3\), then they both are, which is a contradiction. So neither is. If \( k \) is even then \( x^k \) and \( y^k \) are both \( 1 \) mod \( 3 \), so \( x^k+y^k \) is not divisible by any power of \( 3 \).

Now suppose \( k \) is odd. If \(n=0 \) then \( x^k+y^k = 1 \), and there are no solutions to this in positive integers, so we can exclude this case. Since \(n \ge 1 \), \( 3|(x+y) \). Apply LTE:
\[
n = v_3(x^k+y^k) = v_3(x+y) + v_3(k).
\]
So then
\[
x^k + y^k = 3^n = 3^{v_3(x+y)}3^{v_3(k)} = (x+y)k.
\]
The point is that the left side is usually much bigger than the right side, so the result will follow from some routine inequalities.

In that case \( 2^{k-2} > k \) already unless \( k=3,4 \), but \( k \) is odd so \( k = 3\). It's easy to check that \( x=2,y=1,k=3 \) is a solution, as is \( x=1,y=2,k=3 \) if we relax the \( x>y \) assumption, and the above analysis has shown that these are the only ones. \( \square \)

Now write \( a = \frac{x}{z} \) and \( b = \frac{y}{z} \) as quotients of positive integers, with a common denominator. Choose \( z \) as small as possible. Then the conditions of the problem imply that
\[
z^n | (x^n-y^n) \ \text{for all} \ n.
\]
Suppose \( p \) is a prime dividing \( z \). Note \( z|(x-y) \) so \( p|(x-y) \). If \( p|x \) then \( p|y \) as well, but that violates the choice of \( z\): we could write \( a = \frac{x/p}{z/p}, b = \frac{y/p}{z/p} \) to get a smaller common denominator.

So \(p \nmid x,y\) and we are set up to apply LTE. If \( p \) is odd then
\[
n \le v_p(z^n) = v_p(x^n-y^n) = v_p(x-y) + v_p(n).
\]
Taking \( p \) to both sides gives
\[
\begin{align}
p^n &\le (x-y)n \\
\frac{p^n}{n} &\le (x-y)
\end{align}
\]
but this is impossible since the left side goes to infinity as \(n \to \infty\), and the right side is a constant independent of \( n \).

If \( p = 2\) we get
\[
n \le v_2(z^n) = v_2(x^n-y^n) = v_2(x-y)+v_2(n)+v_2(x+y)-1
\]
so
\[
\frac{2^{n+1}}{n} \le (x-y)(x+y)
\]
and we get a similar contradiction.

The conclusion is that there is no prime \( p \) dividing \( z \). So \( z=1 \) and \( a \) and \( b \) are both positive integers. \( \square\)